# Thread: Determine the cycle decomposition of the permutations Originally Posted by Klaas van Aarsen I didn't check them all, but indeed it can't be right that $2$ is in this list twice, nor that it maps to two different elements. Why is it in the list twice?
We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?   Reply With Quote

2.

3. Originally Posted by mathmari We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?
No. When we consider the image of an element such as $2$, we must evaluate all cycles from right to left. We are not supposed to start in the middle.   Reply With Quote

Ok !!

Now I think I got the right results :

$\pi_4=\pi_1\circ\pi_2=(14\ 13) \ (6\ 3\ 2\ 9\ 1)\ (12\ 8\ 7\ 4\ 5)$

$\pi_5=\pi_2\circ\pi_3=(1\ 4) \ (2\ 8\ 11\ 12\ 6\ 7)\ (13\ 14)$

$\pi_6=\pi_2\circ\pi_1=(3 \ 4 \ 9 \ 12\ 1) \ (2\ 5\ 8\ 6\ 7)\ (13 \ 14)$

For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?   Reply With Quote

5. Originally Posted by mathmari For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?
We can write $(a\,b\,c\,d)$ for instance as either $(a\,d)(a\,c)(a\,b)$ or $(a\,b)(b\,c)(c\,d)$. Note that more generally a cycle with an even number of elements can be written as an odd number of 2-cycles (signum -1).
And a cycle with an odd number of elements can be written as an even number of 2-cycles (signum +1).   Reply With Quote

I am looking aagain at the previous question, at the calculation of $\pi_4$. Do we get the same results if we use the initially given $\pi_1$ and $\pi_2$ and the same if we use the cycle decomposition of them? I get the following from the cycle decomposition:

$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

$\pi_4=\pi_1\circ\pi_2=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 ) \ (7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

$14\rightarrow 13$

$13\rightarrow 14$

$6\rightarrow 12\rightarrow 8$

$12\rightarrow 1\rightarrow 9$

$1\rightarrow 6\rightarrow 12$

$7\rightarrow 8\rightarrow 6$

$8\rightarrow 5$

$5\rightarrow 4\rightarrow 2$

$4\rightarrow 2\rightarrow 3$

$2\rightarrow 7$

$9\rightarrow 4$

$3\rightarrow 1$

So $\pi_4=(14 \ 13) \ (6 \ 8 \ 5 \ 2 \ 7) \ (12 \ 9 \ 4 \ 3 \ 1)$.

And from the initially given forms:

$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$

$\pi_4=\pi_1\circ\pi_2=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12) \ (2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$

$13 \rightarrow 14$

$14\rightarrow 13$

$1\rightarrow 12 \rightarrow 6$

$12\rightarrow 6\rightarrow 8$

$6\rightarrow 1 \rightarrow 3$

$2\rightarrow 4\rightarrow 9$

$4\rightarrow 5$

$5\rightarrow 8\rightarrow 12$

$8\rightarrow 7$

$7\rightarrow 2\rightarrow 4$

$3\rightarrow 1\rightarrow 2$

$9\rightarrow 1$

So $\pi_4=(13 \ 14) \ (1 \ 6 \ 3 \ 2 \ 9 ) \ (12 \ 8 \ 7 \ 4 \ 5)$.

Have I done something wrong or is it correct that we get different results?   Reply With Quote

7. The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?   Reply With Quote Originally Posted by Klaas van Aarsen The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it? What do you mean? I got stuck right now.   Reply With Quote

9. Originally Posted by mathmari I get the following from the cycle decomposition:

$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$
You wrote (6 12 1) here. Originally Posted by mathmari And from the initially given forms:

$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 1 4)$
And you wrote (1 12 6) here.
Shouldn't they be the same? Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition?   Reply With Quote Originally Posted by Klaas van Aarsen You wrote (6 12 1) here.

And you wrote (1 12 6) here.
Shouldn't they be the same? Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition? Oh yes, you are right! I used theinverse accidentaly. So we get the same result using the initial permutation and the cycle decomposition of it.  Reply With Quote

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